wu :: forums « wu :: forums - Complex Principal Square Roots (8/26/2002) » Welcome, Guest. Please Login or Register. Jul 6th, 2022, 3:50pm RIDDLES SITE WRITE MATH! Home Help Search Members Login Register wu :: forums  riddles  putnam exam (pure math) (Moderators: Grimbal, william wu, Icarus, towr, SMQ, Eigenray)  Complex Principal Square Roots (8/26/2002) « Previous topic | Next topic » Author Topic: Complex Principal Square Roots (8/26/2002)  (Read 2017 times)
william wu       Gender: Posts: 1291 Complex Principal Square Roots (8/26/2002)   « on: Aug 26th, 2002, 10:10pm » Quote Modify

The complex Principal Square Root sqrt(z) of a complex number z is the square root (one of two if z != 0) whose Real Part is same sign to Im(sqrt(z)) as to Im(z). (Yes, zero has a sign.)

Now let
f(z) := sqrt(1 - z^2), g(z) := sqrt(1-z) * sqrt(1+z)
F(z) := sqrt(z^2 - 1), G(z) := sqrt(z-1) * sqrt(z+1)

Over what region in the complex plane does f(z) = g(z)?
Over what region in the complex plane does F(z) = G(z)?
 « Last Edit: Sep 16th, 2002, 2:00pm by william wu » IP Logged

[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
william wu       Gender: Posts: 1291 Re: Complex Square Roots   « Reply #1 on: Aug 27th, 2002, 7:04pm » Quote Modify

To start, here's how to find the square root of a complex number a + bi:

Let w = c + di be the square root of a + bi. Then:

c + di = sqrt(a + bi)

Squaring both sides:

(c + di)^2 = a + bi
c^2 + 2cdi - d^2 = a + bi

This gives us two equations in terms of c and d:

c^2 - d^2 = a
2cd = b

Solve for c and d to get the square root, c + di. IP Logged

[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
Yournamehere
Guest  Re: Complex Principal Square Roots   « Reply #2 on: Aug 29th, 2002, 2:16pm » Quote Modify Remove

on Aug 26th, 2002, 10:10pm, william wu wrote:
 The complex Principal Square Root sqrt(z) of a complex number z is the square root (one of two if z != 0) whose Real Part is same sign to Im(sqrt(z)) as to Im(z). (Yes, zero has a sign.)

What exactly does "Real Part is same sign to Im(sqrt(z)) as to Im(z)" mean?  Does this mean signum(Im(sqrt(z))) = signum(Im(z)), or does it mean signum(Re(sqrt(z)))*signum(Im(sqrt(z))) = signum(Re(sqrt(z)))*signum(Im(z)), or does it mean something else? IP Logged
william wu       Gender: Posts: 1291 Re: Complex Principal Square Roots   « Reply #3 on: Sep 3rd, 2002, 6:51pm » Quote Modify

Sorry for the wording; I was aware that it was unclear but that's copied verbatim from the blackboard.

I think it means that sign(Re(sqrt(z))) = sign(Im(sqrt(z))) = sign(Im(z)). IP Logged

[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
Yournamehere
Guest  Re: Complex Principal Square Roots   « Reply #4 on: Sep 4th, 2002, 10:25am » Quote Modify Remove

That definition can't be right.   Neither square root of -i, for instance, satisfies signum(Re(sqrt(-i))) = signum(Im(sqrt(-i))).

The straightforward approach is simply to enumerate cases.  Both f and g (and F and G) are both zeros of some polynomial z^2+k.  From the Fundamental Theorem of Algebra, there are only two complex zeros, and it's easy to show that if the zeros are a and b, then, in polar notation, |a| = |b| and Arg(a) = Arg(b)+\pi.  So you only need to determine whether f(z) and g(z) both lie in some half of the Cartesian plane or not .  If they do lie in the same half-plane, then they must be equal;  if not, they must be different.  This is much easier than computing f(z) and g(z) explicitly.

Exactly what the cases are to split on depends on the definition of Principal Square Root.  The usual definition I understood is -\pi/2 < Arg(Sqrt(z)) <= \pi/2, but I cannot make this correspond at all with the definition given in the problem.  I'll post a solution based on this definition later.

 Technically you can't use the usual definition of a half-plane (open or closed);  the plane must include half of its boundary points.  Sort of "half-open, half-closed".  -\pi/2 < Arg(z) <= \pi/2 is an example of such. IP Logged
Yournamehere
Guest  Re: Complex Principal Square Roots   « Reply #5 on: Sep 4th, 2002, 10:07pm » Quote Modify Remove

Looking back at what I wrote this morning, I think I could have been clearer and said "g^2=f^2, so either g=f or g=-f".  Anyways, let's assume Principal Square Root is defined by -\pi/2 < Arg Sqrt(z) <= \pi/2.  Recall that for any two complex numbers s, t, Arg (st) = Arg s + Arg t, and also Arg Sqrt(z) = (Arg z)/2.  Now consider the following cases:

Case 1:  Im z > 0.  Then Im (1+z) > 0, so 0 < Arg (1+z) < \pi, and 0 < Arg Sqrt(1+z) < \pi/2.  Also Im (1-z) < 0, so -\pi < Arg (1-z) < 0, and -\pi/2 < Arg Sqrt(1-z) < 0.  Thus -\pi/2 < Arg (Sqrt(1-z) Sqrt(1+z)) < pi/2.  Since -\pi/2 < Arg f <= \pi/2, g cannot equal -f, so g=f.

Case 2:  Im z < 0.  Similar reasoning shows g=f as well.

Case 3:  Im z = 0.  Easy to show g=f, since now 1-z and 1+z are guaranteed to be real.

So g=f for all z.

Part (b) is a bit trickier.  Note that the above analysis does not give a clear answer, for although 0 < Arg Sqrt(z+1) < \pi/2, we have Im (z-1) > 0, so 0 < Arg Sqrt(z-1) < \pi/2, and hence 0 < Arg (Sqrt(z-1) Sqrt(z+1)) < \pi.  So either 0 < Arg G <= \pi/2 (and thus we must have G=F), or \pi/2 < Arg G < \pi, and G must be -F.  So we need to do something to further distinguish these two outcomes.  I'll only give a hint to the rest, in that it might be difficult to put further bounds on Arg G, and that it might be easier to find other ways to determine if G=F.

I'm not sure how friendly a message board like this is to discussing math problems;  formatting and notation can be difficult to convey. IP Logged
Yournamehere
Guest  Re: Complex Principal Square Roots   « Reply #6 on: Sep 5th, 2002, 10:05am » Quote Modify Remove

on Sep 4th, 2002, 10:07pm, Yournamehere wrote:
 Note that the above analysis does not give a clear answer, for although 0 < Arg Sqrt(z+1) < \pi/2, we have Im (z-1) > 0, so 0 < Arg Sqrt(z-1) < \pi/2, and hence 0 < Arg (Sqrt(z-1) Sqrt(z+1)) < \pi.

Oops.  I should have noted that this refers to "Case 1" only.   Case 2 runs into a similar problem, though. IP Logged

 Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs => putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »